Question

我在Python中有一个np.datetime64日期列表：

['2016-12-01T02:00:00.000000000', '2016-12-01T04:00:00.000000000',
 '2016-12-01T06:00:00.000000000', '2016-12-01T08:00:00.000000000',
 '2016-12-01T10:00:00.000000000', '2016-12-01T12:00:00.000000000', 
 '2016-12-01T14:00:00.000000000', '2016-12-01T16:00:00.000000000', 
 '2016-12-01T18:00:00.000000000', '2016-12-01T20:00:00.000000000', 
 '2016-12-01T22:00:00.000000000', '2016-12-02T00:00:00.000000000', 
 '2016-12-02T02:00:00.000000000', '2016-12-02T04:00:00.000000000', 
 '2016-12-02T06:00:00.000000000', '2016-12-02T08:00:00.000000000', 
 '2016-12-02T10:00:00.000000000', '2016-12-02T12:00:00.000000000', 
 '2016-12-02T14:00:00.000000000', '2016-12-02T16:00:00.000000000', 
 '2016-12-02T18:00:00.000000000', '2016-12-02T20:00:00.000000000', 
 '2016-12-02T22:00:00.000000000', '2016-12-03T00:00:00.000000000', 
 '2016-12-03T02:00:00.000000000', '2016-12-03T04:00:00.000000000',
 '2016-12-03T06:00:00.000000000', '2016-12-03T08:00:00.000000000', 
 '2016-12-03T10:00:00.000000000', '2016-12-03T12:00:00.000000000', 
 '2016-12-03T14:00:00.000000000', '2016-12-03T16:00:00.000000000', 
 '2016-12-03T18:00:00.000000000', '2016-12-03T20:00:00.000000000', 
 '2016-12-03T22:00:00.000000000']

我希望在列表中的每个日历日循环。我试图从列表中提取每个唯一的日期（即找到最小和最大日期并创建这些日期之间的日期列表）但这对我想做的事情并不理想。

我希望的结果是让代码允许我循环遍历列表中的每个日期/日历日并获取与此日期对应的日期时间：

for each_date in date_list:
    ***get all datetimes corresponding to each_date***

(loop would occur 3 times in this example)

注：

1）迭代每个[n：n + 24]或任何不会每天都不起作用的解决方案将具有相同的时间步数。

Answer 1

如果时间戳是有序的，我们可以使用itertools.groupby函数在相应的日期对数组元素进行分组。

可以使用np.datetime64.astype(..., dtype='datetime64[D]')获取日期，因此我们可以将其写为：

from numpy import datetime64
from functools import partial
from itertools import groupby

for day, timestamps in groupby(data_array,
                               partial(datetime64.astype, dtype='datetime64[D]')):
    # process day and timestamps
    pass

此处day是datetime64[D] numpy对象（仅包含当天），timestamps是可迭代（不是列表，但我们可以将其转换为相应时间戳的列表。 data_array是包含初始数据的数组。

例如：

>>> for day, timestamps in groupby(data_array,
...                                partial(datetime64.astype, dtype='datetime64[D]')):
...     print((day, list(timestamps)))
... 
(numpy.datetime64('2016-12-01'), [numpy.datetime64('2016-12-01T02:00:00.000000000'), numpy.datetime64('2016-12-01T04:00:00.000000000'), numpy.datetime64('2016-12-01T06:00:00.000000000'), numpy.datetime64('2016-12-01T08:00:00.000000000'), numpy.datetime64('2016-12-01T10:00:00.000000000'), numpy.datetime64('2016-12-01T12:00:00.000000000'), numpy.datetime64('2016-12-01T14:00:00.000000000'), numpy.datetime64('2016-12-01T16:00:00.000000000'), numpy.datetime64('2016-12-01T18:00:00.000000000'), numpy.datetime64('2016-12-01T20:00:00.000000000'), numpy.datetime64('2016-12-01T22:00:00.000000000')])
(numpy.datetime64('2016-12-02'), [numpy.datetime64('2016-12-02T00:00:00.000000000'), numpy.datetime64('2016-12-02T02:00:00.000000000'), numpy.datetime64('2016-12-02T04:00:00.000000000'), numpy.datetime64('2016-12-02T06:00:00.000000000'), numpy.datetime64('2016-12-02T08:00:00.000000000'), numpy.datetime64('2016-12-02T10:00:00.000000000'), numpy.datetime64('2016-12-02T12:00:00.000000000'), numpy.datetime64('2016-12-02T14:00:00.000000000'), numpy.datetime64('2016-12-02T16:00:00.000000000'), numpy.datetime64('2016-12-02T18:00:00.000000000'), numpy.datetime64('2016-12-02T20:00:00.000000000'), numpy.datetime64('2016-12-02T22:00:00.000000000')])
(numpy.datetime64('2016-12-03'), [numpy.datetime64('2016-12-03T00:00:00.000000000'), numpy.datetime64('2016-12-03T02:00:00.000000000'), numpy.datetime64('2016-12-03T04:00:00.000000000'), numpy.datetime64('2016-12-03T06:00:00.000000000'), numpy.datetime64('2016-12-03T08:00:00.000000000'), numpy.datetime64('2016-12-03T10:00:00.000000000'), numpy.datetime64('2016-12-03T12:00:00.000000000'), numpy.datetime64('2016-12-03T14:00:00.000000000'), numpy.datetime64('2016-12-03T16:00:00.000000000'), numpy.datetime64('2016-12-03T18:00:00.000000000'), numpy.datetime64('2016-12-03T20:00:00.000000000'), numpy.datetime64('2016-12-03T22:00:00.000000000')])

因此，我们每天都选择打印相应timestamps的列表，但这当然是一个选项。与示例显示的不同，并非所有切片都具有相同的长度（最后两个切片具有额外的元素）

请注意timestamps是一个迭代器，因此如果你没有将它转换为一个列表就会耗尽，然后在一个循环之后，迭代器耗尽。

groupby以线性时间工作，因为每次检查＆＃34;组密钥＆＃34;与前一个元素相同，但如前所述，必须对数据进行排序。

Answer 2

您可以将collections.defaultdict用于O（n）解决方案。您可以使用Pandas来规范化datetime个对象，尽管这也可以通过NumPy实现。

import pandas as pd
from collections import defaultdict

d = defaultdict(list)

for item in L:
    day = pd.to_datetime(item).normalize().to_datetime64()
    d[day].append(item)

print(d)

defaultdict(list,
            {numpy.datetime64('2016-12-01T00:00:00.000000000'):
                 [numpy.datetime64('2016-12-01T02:00:00.000000000'),
                  ...
                  numpy.datetime64('2016-12-01T22:00:00.000000000')],
             numpy.datetime64('2016-12-02T00:00:00.000000000'):
                 [numpy.datetime64('2016-12-02T00:00:00.000000000'),
                  ...
                  numpy.datetime64('2016-12-02T22:00:00.000000000')],
             numpy.datetime64('2016-12-03T00:00:00.000000000'):
                 [numpy.datetime64('2016-12-03T00:00:00.000000000'),
                  ...
                  numpy.datetime64('2016-12-03T22:00:00.000000000')]})

使用Python中的日期列表循环24小时

2 个答案: